package cn.yuan.arithmetic.sort;

/**
 * 并归排序实现.
 *
 * @author : yjs
 * @date: 2020-12-17 11:16
 */
public class MergeSort {
    public static void main(String[] args) {
        int[] array = {3, 4, 7, 4, 1, 9, 13, 24, 11, 45, 22, 45};
        System.out.println(print(array));
        MergeSort.sort(array, 0, array.length - 1);
        System.out.println(print(array));
    }

    /**
     * 归并排序 简介:将两个（或两个以上）有序表合并成一个新的有序表
     * 即把待排序序列分为若干个子序列，每个子序列是有序的。然后再把有序子序列合并为整体有序序列 时间复杂度为O(nlogn) 稳定排序方式
     *
     * @param nums 待排序数组
     * @return 输出有序数组
     */
    public static int[] sort(int[] nums, int low, int high) {
        int mid = (low + high) / 2;
        if (low < high) {
            // 左边
            sort(nums, low, mid);
            // 右边
            sort(nums, mid + 1, high);
            // 左右归并
            merge(nums, low, mid, high);
        }
        return nums;
    }

    public static void merge(int[] nums, int low, int mid, int high) {
        int[] temp = new int[high - low + 1];
        int i = low;// 左指针
        int j = mid + 1;// 右指针
        int k = 0;
        // 把较小的数先移到新数组中
        while (i <= mid && j <= high) {
            if (nums[i] < nums[j]) {
                temp[k++] = nums[i++];
            } else {
                temp[k++] = nums[j++];
            }
        }
        // 把左边剩余的数移入数组
        while (i <= mid) {
            temp[k++] = nums[i++];
        }
        // 把右边边剩余的数移入数组
        while (j <= high) {
            temp[k++] = nums[j++];
        }
        // 把新数组中的数覆盖nums数组
        for (int k2 = 0; k2 < temp.length; k2++) {
            nums[k2 + low] = temp[k2];
        }
    }

    /**
     * 打印
     */
    private static String print(int[] array) {
        StringBuilder sb = new StringBuilder("[");
        for (int i = 0; i < array.length; i++) {
            sb.append(array[i]);
            if (i < array.length - 1) {
                sb.append(",");
            }
        }
        sb.append("]");
        return sb.toString();
    }
}
